Hydrostatic force
- Pressure at depth
- p = ρgh
- Force on a vertical wall
- F = ρg·hc·A
- Centre of pressure
- yp = yc + Ic/(ycA)
- Rectangle
- Ic = bH³/12
Hydraulic Engineering: cram pack
Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.
1Formulas and methods
Continuity & Bernoulli
- Continuity
- A1v1 = A2v2
- Bernoulli
- p1 + ½ρv1² + ρgz1 = p2 + ½ρv2² + ρgz2
- Flow rate
- Q = A·v
- Circle area
- A = πd²/4
Pipe friction & head loss
- Reynolds number
- Re = V·D/ν = ρ·V·D/μ
- Flow regime
- laminar below Re ≈ 2000, turbulent above Re ≈ 4000
- Mean velocity
- V = Q/A, A = π·D²/4
- Darcy–Weisbach
- hf = f·(L/D)·V²/(2g)
- Minor losses
- hm = ΣK·V²/(2g)
- Typical K values
- entrance 0.5, exit 1.0, 90° bend ≈ 0.3
- Pressure drop
- Δp = ρ·g·hf
- Two open reservoirs
- Δz = (f·L/D + ΣK)·V²/(2g)
- Free outlet
- H = (1 + f·L/D + ΣK)·V²/(2g)
Pumps & pump power
- Hydraulic power
- Ph = ρ·g·Q·H = Q·Δp
- Shaft power
- Pshaft = Ph/ηpump
- Electrical power
- Pel = Pshaft/ηmotor
- Pump head
- H = Δz + Δp/(ρ·g) + hL
- System losses
- hL = (f·L/D + ΣK)·V²/(2g)
Open-channel flow (Manning)
- Manning equation
- V = (1/n)·R2/3·S1/2
- Flow rate
- Q = A·V
- Hydraulic radius
- R = A/P (P = wetted perimeter, free surface excluded)
- Rectangle
- A = b·y, P = b + 2y
- Trapezoid (side slope z:1)
- A = (b + z·y)·y, P = b + 2y·√(1 + z²)
- Circular pipe, full or half full
- R = D/4
- Required slope
- S = (n·Q/(A·R2/3))²
2Practice problems
1.Hydrostatic forceMedium
A vertical rectangular gate 1 m wide sits in the wall of a tank of water, with its top edge 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 mH = 1.5 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 3 m wide sits in the wall of a tank of oil, with its top edge 1.5 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1.5 mH = 2 mb = 3 mρ = 880 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 150 mm to 40 mm. In the wide section the pressure is 340 kPa and the velocity is 1 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 150 mmd2 = 40 mmp1 = 340 kPav1 = 1 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
A horizontal venturi meter (inlet 80 mm, throat 40 mm) carries water. A mercury U-tube manometer connected between the inlet and the throat shows a level difference of 300 mm. The discharge coefficient is 0.97. Find the flow rate Q. Use ρ = 1000 kg/m³ for water, 13 600 kg/m³ for mercury and g = 9.81 m/s².
d1 = 80 mmd2 = 40 mmΔh = 300 mmCd = 0.97ρHg = 13 600 kg/m³g = 9.81 m/s²
Q = L/s
5.Pipe friction & head lossMedium
Water flows at 2 L/s through a pipe 100 m long with an inside diameter of 50 mm. Use the Darcy friction factor f = 0.025 and g = 9.81 m/s². Find the friction head loss hf.
Q = 2 L/sL = 100 mD = 50 mmf = 0.025g = 9.81 m/s²
hf = m
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 150 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.022. The water surface of the upper reservoir is 30 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 3 bends (K = 0.3 each), a valve (K = 10) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². Find the flow rate Q in L/s.
Δz = 30 mL = 150 mD = 200 mmf = 0.022Kentrance = 0.5Kbend = 0.3 (×3)Kvalve = 10Kexit = 1.0
Q = L/s
7.Pumps & pump powerMedium
In a test, a pump delivers 50 L/s of water (ρ = 1000 kg/m³) against a head of 68 m while its shaft takes 44.5 kW. Use g = 9.81 m/s². Find the pump efficiency η in percent.
Q = 50 L/sH = 68 mPshaft = 44.5 kWg = 9.81 m/s²
η = %
8.Pumps & pump powerHard
A pump moves 40 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 7 m higher. The pipe is 100 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.02; the minor-loss coefficients add up to ΣK = 2.5 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
Q = 40 L/sΔz = 7 mL = 100 mD = 150 mmf = 0.02ΣK = 2.5
H = m
9.Open-channel flow (Manning)Medium
A trapezoidal clean earth channel has a bottom width of 5 m and side slopes of 3:1 (horizontal:vertical). Water flows uniformly 1.2 m deep. Manning n = 0.022 and the bed slope is S = 0.003. Find the flow rate Q in m³/s.
b = 5 my = 1.2 mz = 3n = 0.022S = 0.003
Q = m³/s
10.Open-channel flow (Manning)Hard
A concrete sewer with an inside diameter of 600 mm (Manning n = 0.013) must reach a self-cleansing velocity of 0.9 m/s when flowing just full. Find the minimum slope S (in m/m).
D = 600 mmn = 0.013V = 0.9 m/s
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 33.11 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Centroid depth and area
yc = d + H/2 = 1.5 + 1.5/2 = 2.25 m
A = H·b = 1.5·1 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.25·1.5 = 33.11 kN
Answer
F = 33.11 kN
2.Hydrostatic force
Answer: F = 129.5 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Centroid depth and area
yc = d + H/2 = 1.5 + 2/2 = 2.5 m
A = H·b = 2·3 = 6 m²
Resultant force
F = ρ·g·yc·A = 880·9.81·2.5·6 = 129.5 kN
Answer
F = 129.5 kN
3.Continuity & Bernoulli
Answer: p2 = 241.6 kPa
Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.
Continuity and Bernoulli
Work in Pa, then divide by 1000 to get kPa.
v2 = v1·(d1/d2)² = 1·(150/40)² = 14.06 m/s
½ρ(v1² − v2²) = ½·1000·(1² − 14.06²)/1000 = −98.38 kPa
p2 = p1 + ½ρ(v1² − v2²) = 340 + (−98.38) = 241.6 kPa
Answer
p2 = 241.6 kPa
4.Continuity & Bernoulli
Answer: Q = 10.84 L/s
Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.
Pressure difference from the manometer
Δp = (13600 − 1000)·9.81·0.3 = 37082 Pa
Throat velocity
Continuity and Bernoulli together give the ideal throat velocity.
β = 40/80 = 0.5
v2 = √(2·37082/(1000·(1 − 0.0625))) = 8.894 m/s
Actual flow rate
Q = Cd·A2·v2 = 0.97·0.001257·8.894 = 0.01084 m³/s = 10.84 L/s
Answer
Q = 10.84 L/s
5.Pipe friction & head loss
Answer: hf = 2.644 m
Pipe flow: the Reynolds number Re = V·D/ν = ρ·V·D/μ tells laminar flow (Re below about 2000) from turbulent flow (above about 4000). Friction along a pipe costs the Darcy–Weisbach head loss hf = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses hm = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = hf + hm, with the exit loss K = 1 included.
Mean velocity
A = π·0.05²/4 = 0.001963 m²
V = Q/A = 0.002/0.001963 = 1.019 m/s
Velocity head
V²/(2g) = 1.019²/(2·9.81) = 0.05288 m
Darcy–Weisbach
hf = f·(L/D)·V²/(2g) = 0.025·(100/0.05)·1.019²/(2·9.81) = 2.644 m
Answer
hf = 2.644 m
6.Pipe friction & head loss
Answer: Q = 141.8 L/s
Pipe flow: the Reynolds number Re = V·D/ν = ρ·V·D/μ tells laminar flow (Re below about 2000) from turbulent flow (above about 4000). Friction along a pipe costs the Darcy–Weisbach head loss hf = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses hm = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = hf + hm, with the exit loss K = 1 included.
Loss coefficients
f·L/D = 0.022·150/0.2 = 16.5
ΣK = 0.5 + 3·0.3 + 10 + 1.0 = 12.4
Energy equation
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
30 = (16.5 + 12.4)·V²/(2·9.81)
V = √(2·9.81·30/28.9) = 4.513 m/s
Flow rate
Q = V·π·D²/4 = 4.513·0.03142 = 0.1418 m³/s = 141.8 L/s
Answer
Q = 141.8 L/s
7.Pumps & pump power
Answer: η = 74.95 %
Pump power: the power given to the water is the hydraulic power Ph = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: Pshaft = Ph/ηpump, and the motor draws Pel = Pshaft/ηmotor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.
Hydraulic power
Ph = ρ·g·Q·H = 1000·9.81·0.05·68 = 33.35 kW
Efficiency
η = Ph/Pshaft = 33.35/44.5 = 74.95%
Answer
η = 74.95%
8.Pumps & pump power
Answer: H = 11.13 m
Pump power: the power given to the water is the hydraulic power Ph = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: Pshaft = Ph/ηpump, and the motor draws Pel = Pshaft/ηmotor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.
Velocity
V = Q/A = 0.04/0.01767 = 2.264 m/s
V²/(2g) = 0.2611 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (13.33 + 2.5)·0.2611 = 4.135 m
Pump head
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
H = Δz + hL = 7 + 4.135 = 11.13 m
Answer
H = 11.13 m
9.Open-channel flow (Manning)
Answer: Q = 22.5 m³/s
Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R2/3·S1/2 and Q = A·V, with n the roughness coefficient, S the bed slope (m/m) and R = A/P the hydraulic radius: flow area divided by wetted perimeter. Only the boundary in contact with the water counts in P, not the free surface. Rectangle: A = b·y, P = b + 2y. Trapezoid with side slopes z:1 (horizontal:vertical): A = (b + z·y)·y, P = b + 2y·√(1 + z²). A circular pipe flowing full or half full has R = D/4.
Geometry
A = (b + z·y)·y = (5 + 3·1.2)·1.2 = 10.32 m²
P = b + 2y·√(1 + z²) = 5 + 2·1.2·√(1 + 3²) = 12.59 m
R = A/P = 0.8197 m
Velocity
V = (1/0.022)·0.81972/3·0.0031/2 = 2.181 m/s
Flow rate
Q = A·V = 10.32·2.181 = 22.5 m³/s
Answer
Q = 22.5 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.001718 m/m
Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R2/3·S1/2 and Q = A·V, with n the roughness coefficient, S the bed slope (m/m) and R = A/P the hydraulic radius: flow area divided by wetted perimeter. Only the boundary in contact with the water counts in P, not the free surface. Rectangle: A = b·y, P = b + 2y. Trapezoid with side slopes z:1 (horizontal:vertical): A = (b + z·y)·y, P = b + 2y·√(1 + z²). A circular pipe flowing full or half full has R = D/4.
Hydraulic radius
R = D/4 = 0.6/4 = 0.15 m
Solve Manning for S
S = (n·V/R2/3)² = (0.013·0.9/0.2823)² = 0.001718
Answer
S = 0.001718
That is a fall of 1.718 m per kilometre.
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